# The first postulate of relativity

1. Dec 14, 2012

### metalrose

I just want to make sure that the following assumption is equivaent to the first postulate.

Consider two inertial frames S and S'. By inertial frame, is meant, a frame where all the laws of physics are valid in their simplest form.

Now let an observer in S measure the velocity of S' to be 'v' towards his/her right.
Now we automatically assume, that when an observer in S' measures the velocity of S, he/she will measure it to be 'v' to his/her left.

How can we say that the two observers will agree on the magnitude of the velocities they measure, of the other frame, from their frame?

Is this just an assumption equivalent to the first postulate of relativity which says that :
Given an inertial frame S, a frame S' moving with constant velocity with respect to S, will also be inertial.

According o the above postilate then,

Given S is inertial, and an observer in S measures constant v as the velocity of S', this implies S' is inertial.

Now we know S' is inertial and S is also inertial, so an observer in S' measuring the velocity of S would observe a constant velocity according to the first postulate. It doesn't however garauntee that this velocity's magnitude is going to be v itself.

2. Dec 14, 2012

### phinds

That seems to be like saying that if you put a minus sign in front of an integer, you have changed the absolute value of the integer. Just doesn't seem to make sense.

3. Dec 14, 2012

### ghwellsjr

I think you are just pointing out why the first postulate (the principle of relativity) by itself is not enough to establish Einstein's Theory of Special Relativity which also needs the second postulate (that all light propagates at c in all inertial frames).

4. Dec 14, 2012

### Staff: Mentor

If I understand correctly, you're asking whether SR, like Galilean relativity before it, assumes that A's speed relative to B is equal to B's speed relative to A (and note that I said "speed" not "velocity"); or whether instead that follows from the first principle?

As an assumption, this is so totally unobjectionable (no observed violations, ever) that I doubt many people spend much time considering the question. But if you want to prove it, there are some pretty convincing reductio ad absurdum arguments.

5. Dec 14, 2012

### metalrose

I'd really like to read upon such arguments. Could you point me to some such arguments??

6. Dec 14, 2012

### Staff: Mentor

Let's assume the opposite. Suppose an observer in S measures the velocity of S' to be v to the right but an observer in S' measures the velocity of S to be u to the left where v≠u. Then, suppose an observer in S constructs a standard gun which fires a standard projectile at a muzzle velocity of v. Then, an observer with an identical gun and projectile could determine if they are in S or S' by measuring the muzzle velocity. That would require that the laws of physics were different in S and S', which violates the principle of relativity. Therefore, the assumption is not compatible with the principle of relativity.

7. Dec 15, 2012

### metalrose

I didn't get the logic.
By muzzle velocity, i understand the velocity of the bullet with respect to the gun kept at rest in a particular frame, just at the moment the bullet leaves the gun.
This velocity will be fixed for a particular make of the gun.

Now an observer in S could fire the gun, and measure the muzzle velocity to be v_m, say.
Now, an observer in S' , with an identical gun, may obsetve the same consequences.
But this has no bearing on the velocities of the others' frame that each of them observes ...

8. Dec 15, 2012

### Staff: Mentor

Not if the principle of relativity doesn't hold. Then it will be fixed for a particular make and reference frame.

9. Dec 15, 2012

### metalrose

Ok, i get that.
To say that the principle of relativity doesn't hold is the same as saying that the laws of physic are differnt in the two frames.
That would lead to the gun behaving differently in the two frames.
Which means the two observers will measure different muzzle velocities.

But this still doesn't lead to the my initial problem of v≠u.

10. Dec 15, 2012

### Fredrik

Staff Emeritus
The first postulate is the principle of relativity, which says roughly that all inertial frames are equivalent. Since that's a rather loosely stated idea, you can't prove rigorously that it's equivalent to (or implies) some other idea. But I would still consider the idea that you're describing to be an aspect of the principle of relativity, since this is something we would expect to hold, if the principle of relativity holds.

Consider e.g. two identical objects A and B attached to a compressed spring that's prevented from expanding e.g. by a string. If you cut the string, the two objects will be shot away from each other. These objects are being subjected to essentially the same thing, so it would be pretty weird if the speed of A relative to B is not the same as the speed of B relative to A.

If those speeds are different, then what would be the reason for that difference? Since the objects are identical, something else would have to differ. For example, there could be something fundamentally different about the two positions in space where the objects started, or the two objects could have started out with a non-zero velocity relative to some kind of invisible "stuff" that fills up all of space, and the speed of one of the objects relative to the "stuff" increased while the speed of the other relative to the "stuff" decreased. These are the sort of things that the principle of relativity is used to rule out.

11. Dec 15, 2012

### metalrose

Lets attach frame S to the object A and frame S' to the object B.
Now your argument definitely makes sense because theres a spring thats doing the same thing to both A and B, and so, unless theres some weird stuff going on related to initial positions and stuff, by symmetry, the relative velocities of A wrt B and vice versa should be the same.

But this entire argument depends upon the common spring, doing the same thing to the two frames.

What if there isn't a spring? Just frame S and S' , having constant relative velocities w.r.t. Each other? How do i now prove that the velocity of S w.r.t. S' is the same as the velocity of S' w.r.t. S ??

12. Dec 15, 2012

### Fredrik

Staff Emeritus
"Prove" is the wrong word in my opinion, since we're not dealing with mathematical statements. This is just a matter of whether a mathematical statement should be thought of as making an aspect of a loosely stated idea mathematically precise. Because of that, I don't think we even need an argument. I would say that if the velocity of S' in S is v, and the velocity of S in S' isn't -v, then that is by itself a good enough reason to not consider S and S' equivalent as far as physics is concerned. If two observers describe each other more differently than this, why would we call them equivalent?

13. Dec 15, 2012

### Studiot

That's a strange definition, don't you find it a bit woolly?

I have seen

An inertial frame is one which is not accelerating.

and

An intertial frame is one that is either at rest or moving with constant velocity.

14. Dec 16, 2012

### metalrose

That makes sense i guess. Thanks a lot.

15. Dec 16, 2012

### metalrose

Accelerating with respect to what ? Rest or constant velocity with respect to what ?

The definition i used is a textbook definition.
An inertial frame is one where all laws of physics are valid.
Now, if some other frame i moving with respect to this inertial frame at constant velocity, then even that becomes an inertial frame.

16. Dec 16, 2012

### strangerep

Acceleration is detectable locally.
No, an inertial frame is one in which the observer feels no acceleration.
See below...
Similarly, an (Einsteinian) inertial frame is a reference frame in which spatial relations, as determined by rigid scales at rest in the frame, are Euclidean and in which there exists a universal time in terms of which free particles remain at rest or continue to move with constant speed along straight lines (i.e., in terms of which free particles obey Newton's first law).

With these definitions, we can formulate (still following Rindler):

Postulate 1 (Principle of Relativity):
The laws of physics are identical in all inertial frames, or, equivalently, the outcome of any physical experiment is the same when performed with identical initial conditions relative to any inertial frame.

Last edited: Dec 16, 2012
17. Dec 16, 2012

### strangerep

FYI, I just generalized some of my calculations and verified that from the requirements of inertial motion, spatial isotropy, and the assumption that boosts along a given axis form a 1-parameter Lie group, one can derive the result that a boost $\Lambda(v)$ has inverse given by $\Lambda(-v)$.

I.e., one does not need to assume that the inverse corresponds to the parameter $-v$, (as is done in some group theoretic derivations).

Afaict, this establishes the frame reciprocity mentioned above.

18. Dec 16, 2012

### metalrose

None of the above made any sense to me, thats because im still a junior in undergrad.
Any resources i could use at this point to understand the above ??

19. Dec 16, 2012

### metalrose

Also,
I quoted my definition of the first postulate and inertial frames from the modern physics textbook by john r. Taylor.

I quote again,

Inertial frame: an inertial frame is any reference frame ( that is, system of coordinates x,y,z,t), where all the laws of physics hold in their simplest form.

I think what you describe is equivalent to above. I.e. spatial relations being euclidian and newton's first law , are some of the laws of physics in their simplest form.

20. Dec 16, 2012

### metalrose

Also, my textbook, just after defining inertial frames in the above way, immediately says,

I think this is equivalent o the definition you gave.

21. Dec 16, 2012

### Fredrik

Staff Emeritus
In my opinion, you don't need to understand the mathematical things that strangerep mentioned to understand the issue you asked about. He mentioned them to me because we've been discussing ways to "discover" the Lorentz group in these threads: 1, 2.

One of the things he mentioned is however directly relevant to what you've been asking about: Isotropy. In the 1+1-dimensional case, we can view the statement that you're concerned about as a precise statement of the idea that "space is the same in all directions". I should probably have mentioned that myself.

To do this, you will at some point have to write down a mathematical statement that you view as the assumption that makes the idea of isotropy mathematically precise. So I'm curious about what that statement is. In my approach for the 1+1-dimensional case, $\Lambda(v)^{-1}=\Lambda(-v)$ is that statement. It can be viewed as the precise meaning of "isotropy" in 1+1 dimensions. In 3+1 dimensions, I think the mathematically precise statement of isotropy should be that the group of proper and orthochronous inertial coordinate transformations has a subgroup that consists of transformations of the form
\begin{pmatrix}1 & 0^T\\ 0 & R\end{pmatrix} where 0 is the 3×1 matrix with all components =0, and R is a 3×3 rotation matrix. This is at least one aspect of isotropy. I still haven't finished typing up my notes on the 1+1-dimensional case, so I haven't had time to think about what my complete list of assumptions will be in the 3+1-dimensional case. In particular, I haven't thought about what exactly should replace $\Lambda(v)^{-1}=\Lambda(-v)$. The statement about rotations above will be a part of it, but I think I will probably also have to assume that boosts satisfy $\Lambda(v)^{-1}=\Lambda(-v)$. I haven't thought this through enough to know if this rule can be derived from other mathematical assumptions.

Last edited: Dec 16, 2012
22. Dec 16, 2012

### Staff: Mentor

Sure it does. Think a little more about how you make any measurement. Any measurement is a comparison of your measured quantity against some standard quantity. So, how do you measure velocity? As with any measurement you compare it to a standard.

One way to do that is to produce standard distances and standard times and determine how many standard distances are traveled in one standard time. Another way is to produce a standard velocity. If your velocity standard and your distance/time standards agree in one reference frame and they disagree in another reference frame then the laws of physics are different in the different frames, which violates the principle of relativity.

So, if you equip two observers with a velocity standard calibrated to v and observer A sees the speed of B as v with the distance/time standards while observer B sees the speed of A as u with the distance/time standards, then in A's frame the velocity standard will match the time/distance standard and in B's frame the velocity standard will not match. This violates the principle of relativity.

23. Dec 16, 2012

### Fredrik

Staff Emeritus
I will elaborate a bit on DaleSpam's gun argument, to make the use of isotropy explicit.

Suppose that S and S' are both inertial coordinate systems such that S' can be obtained from S by a pure boost in the positive x direction. Denote the speed of S' in S by v, and the speed of S in S' by u. Suppose that a gun G is built at rest in S according to specifications that ensure that its bullets will come out with speed v relative to the gun. Suppose that a gun G' is built at rest in S' according to the same specifications. The principle of relativity strongly suggests that the bullets coming out of this gun will have speed v relative to the gun. Now suppose that the gun at rest in S' is rotated around the z axis by an angle of pi (so that it now points in the negative x' direction instead of the positive x' direction). If space is isotropic, the bullets coming out of the gun should still come out at speed v.

Now suppose that both guns are fired at the event where the fronts of the barrells pass each other. The bullet that comes out of G will be comoving with G'. Isotropy strongly suggests that the bullet that comes out of G' will be comoving with G. So the speed of G with respect to the bullet that came out of it (by assumption =u), is the same as the speed of the bullet that came out of G' with respect to G' (by isotropy and the principle of relativity =v). In other words, u=v.

24. Dec 16, 2012

### metalrose

Could you expand a bit more on what you mean by isotropy of space and how the above statement should therefore be true?

25. Dec 16, 2012

### Fredrik

Staff Emeritus
The assumption that space is isotropic means roughly that space is the same in every direction. Every theory that assumes that space is isotropic or proves it as a theorem does it in a different way.

Since we're not working within the framework of a specific theory, there's no exact definition that we can use. We're just talking about how to apply some loosely stated ideas. I'm just saying that if we build a gun and it turns out that the speed of the bullet that comes out of it depends on the direction we aim, then I would interpret that as a violation of isotropy.